Consider the cell:
Pt H2(1atm)IHA2(1M) II HA1(1M)IH2(1 atm)Pt
Given that PKa1 of HA1 is = 3 (at Cathode)
Pka2
for HA2 is = 5 (at anode)
Related Questions:
Showing posts with label ELECTROCHEMISTRY. Show all posts
Showing posts with label ELECTROCHEMISTRY. Show all posts
Friday, May 7, 2021
Two weak acid solution HA1 and HA2 each with same concentration and having Pka values 3 and 5 are placed in contact with hydrogen electrode ( 1 atm at 25 degree) and are interconnected through a salt bridge find the emf.
A current of 3 ampere passed for 2 hours through a solution of CuSO4 3 gm of Cu2+ ions were discharged at cathode calculate the current efficiency.
Calculate the emf of the cell: Pt H2(1atm)ICH3COOH(0.1M) II NH4OH(0.01M)IH2(1 atm)Pt and Ka for CH3COOH= 1.8x10^-5 and Kb for NH4OH = 1.8x10^-5
(A) Reaction at anode:
(B) Reaction at Cathode:
From equation (1)
Alternative Method:
(A) Reaction at anode:
(B) Reaction at Cathode:
From equation (1)
Related Questions:
Thursday, May 6, 2021
How many grams of silver could be plated out on a serving tray by electrolysis of solution containing silver in +1 oxidation state for a period of 8 hours. At a current of 8.46 ampere? What is the area of tray if the thickness of silver plating is 0.00254 cm. density of silver is 10.5 gm cc.
A silver electrode is immersed in saturated solution of Ag2SO4(aq). The potential difference between the silver and standard electrode is found to be 0.711 V. determined Ksp(Ag2SO4) given E0Ag+/Ag= 0.799 V
Tuesday, June 9, 2020
Write correct Nernst equation for the given galvanic cell containing both the electrodes are M/M (insoluble) salt at 25°c.: Ag(s)|AgCl(s)| KCl(M) || KBr (M)|AgBr(s)|Ag(s): [Given Ksp(AgCl) and Ksp(AgCl)]
Important thing :
For concentrated cell both electrodes are same so E not cell is always zero and Qc is always less than one.
You should first assume any one is the cathode and the other one is the anode. Then you can calculate the theoretical emf:
(1) If the emf > 0, then that reaction will be spontaneous and it means you got the right cathode and right anode.
(2) If the calculated emf < 0, it means that that reaction is not spontaneous, which means that the reverse reaction is spontaneous; that happens if you exchange the cathode and anode.
For the galvanic cell: Ag|AgCl(s)| KCl (0.2M) || KBr (0.001M) |AgBr(s) |Ag : calculate the e.m.f. generated and assign correct polarity to each electrode for a spontaneous process after taking an account of cell reaction at 25°c. [Given Ksp(AgCl) = 2.8×10^-10 and Ksp(AgCl) = 3.3×10^-13]
Ag(s)|AgBr(s)| KBr(0.002M) || KCl (0.2M)|AgCl(s)|Ag(s):
Where
Ag(s)|AgBr(s)| KBr(0.002M) act as Anode while KCl (0.2M)|AgCl(s)|Ag(s): act as cathode
How to determine anode and cathode of concentration cell instead of both electrodes are same ?
Important thing :
For concentrated cell both electrodes are same so E not cell is always zero and Qc is always less than one.
You should first assume any one is the cathode and the other one is the anode. Then you can calculate the theoretical emf:
(1) If the emf > 0, then that reaction will be spontaneous and it means you got the right cathode and right anode.
(2) If the calculated emf < 0, it means that that reaction is not spontaneous, which means that the reverse reaction is spontaneous; that happens if you exchange the cathode and anode
Sunday, April 19, 2020
In the electrolysis of aqueous solution of NaOH , 2.8 litre of oxygen at NTP liberated at anode. How much Hydrogen was liberated at cathode ?
We can calculate volume of hydrogen by applying first law of faraday's:
On passing 3 faradays of electricity through three electrolytic cells connect in series containing Ag+ , Ca+2 and Al+3 ions respectively , the molar ratio in which three metals ions are liberated at the electrode is
We are find molar ratio of metal ions by applying faraday's first law of electrolysis
Subscribe to:
Posts (Atom)